# Diff Eq tired spring

1. Oct 11, 2007

### andre_4

1. The problem statement, all variables and given/known data
Determine the minimum positive value of the constant k needed to ensure that the function y(x) obeying

x^2 y'' + K y = 0, y(1) = 0, y'(1)=1

will not simply keep on increasing toward large x but will instead return to cross y=o ocassionally

3. The attempt at a solution
So far I've tried doing a change of variable and trnasform t= ln(X). The idea is that I should get a constant coefficient equation but do not know how to get there yet.
I have tried using the chain rule twice to get from d^2y/dx^2 to d^2y/dt^2 but as said I do not seem to make any progress. Any ideas please?

2. Oct 11, 2007

### rock.freak667

Use the trial solution $$y=x^m$$

3. Oct 12, 2007

### HallsofIvy

Staff Emeritus
That's an "Euler type" equation. As rock.freak667 said, you can reduce to the "characteristic equation" by trying y= xm. However, I suspect that equation will have complex solutions and it's not clear how to reduce that to real solutions to the d.e.

Yes, you can always reduce an "Euler type" equation to a constant coefficients equation by the substitution t= ln(x). You use the chain rule to change:
$$\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{t}\frac{dy}{dt}$$
$$\frac{d^2y}{dx^2}= \frac{d }{dx}\left(\frac{1}{t}\frac{dy}{dt}\right)$$
$$= \frac{1}{t}\frac{d }{dt}\left(\frac{1}{t}\frac{dy}{dt}\right)$$
In fact, the characteristic equation for the constant coefficients equation you get this way is exactly the same as the equation you get by trying y= xm but now you know that t= ln(x) so you can see what to do with complex solutions- exactly what you would do for a constant coefficients equation and then replace t by ln(x).

Last edited: Oct 12, 2007